How to print Nth column of tabular data using bash only

Question

I have a variable like

date="January 01 2019"

I want to split it into different variables like

month=$(echo $date|awk '{print $1}')

only I want to simplify it by doing this operation just using bash functionality. Is there any way to do it? I can tell bash to only print, say, the first 3 characters of the variable

month=${date: 0: 7}

which would not work for other months.
Is there any way to do this?

try: `read month day year <<<"$date"`, then `echo $year $month $day`... — F. Hauri - Give Up GitHub, Oct 17 '19 at 12:15
thanks @F.Hauri it is bash functionality but I'm looking for something short that fits in one short line — Tomas Hrubovcak, Oct 17 '19 at 12:28

RavinderSingh13 · Accepted Answer · 2019-10-17T12:10:52.617

You could try following. To get 1st part of your variable.

echo ${date%% *}

From man bash page:

${parameter%word} ${parameter%%word} Remove matching suffix pattern. The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the %'' case) or the longest matching pattern (the%%'' case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.

nice. can you please explain the logic behind this? how can it be applied on printing other columns? — Tomas Hrubovcak, Oct 17 '19 at 12:08
@TomasHrubovcak, Your welcome, this is called parameter expansion. `%` is for shortest match and `%%` is for the longest match to be deleted from given pattern. — RavinderSingh13, Oct 17 '19 at 12:11

Akhil · Answer 2 · 2019-10-17T12:43:59.323

1

Yes. It's possible. If you are sure there are no leading spaces.

$ date="January 01 2019"
$ echo ${date/%\ */}
January

manual

' \ * - means replace space01 2019 with nothing and print.

edited Oct 17 '19 at 12:43

answered Oct 17 '19 at 12:06

Akhil

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appreciated, but the answer with the green tick works better plus explains why – Tomas Hrubovcak Oct 17 '19 at 12:36

How to print Nth column of tabular data using bash only

2 Answers2